Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 9th
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In the Figure, ABCD is a rectangle and EFGH is a parallelogram. Using the measurements given in the figure, what is the length d of the segment that is perpendicular to HE¯ and FG¯? - Mathematics

Sum

In the Figure, ABCD is a rectangle and EFGH is a parallelogram. Using the measurements given in the figure, what is the length d of the segment that is perpendicular to `bar("HE")` and `bar("FG")`?

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Solution

In the given figure ABCD is a rectangle and EFGH is a parallelogram.

In the right triangle AEH

HE = `sqrt("AH"^2 + "AE"^2)`

= `sqrt(3^2 + 4^2)`

= `sqrt(9 + 16)`

= `sqrt(25)`

HE = 5

∴ GF = 5 ...(HE and Gf are opposite sides of a parallelogram)

In the right triangle

GC = `sqrt("GF"^2 - "FC"^2)`

= `sqrt(5^2 - 3^2)`

= `sqrt(25 - 9)`

= `sqrt(16)`

∴ DG = 10 – 6 = 4

Area of ΔAEH + Area of ΔBEF + Area of ΔFCG + Area of ΔHDG

= `1/2 xx 3 xx 4 + 1/2 xx 6 xx 5 + 1/2 xx 3 xx 4 + 1/2 xx 5 xx 6`

= (6 + 15 + 6 + 15)

= 42

∴ Area of 4 triangles = 42

Area of the parallelogram = Area of the rectangle ABCD – Area of 4 triangles.

= 10 × 8 – 42

= 80 – 42

= 38

b × h = 38

5 × d = 38

d = `38/5`

= `7 3/5`

Length of d = `7 3/5` or 7.6

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APPEARS IN

Samacheer Kalvi Mathematics Class 9th Tamil Nadu State Board
Chapter 4 Geometry
Exercise 4.2 | Q 10 | Page 158
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